Editing Openai/6908ee36-d3b4-8010-a468-28e0753d08ae (section)

=== Assistant: Nice question — here’s a clean apples-to-apples comparison. ===
Nice question — here’s a clean apples-to-apples comparison.

===== Compton period (hydrogen atom of mass mHm_HmH): =====

TC=hmHc2.T_C=\frac{h}{m_H c^2}.TC=mHc2h.
Your HU “state latency” time:

ΔtHU=α TC,α=0.00843043657.\Delta t_{\text{HU}} = \alpha\,T_C,\quad \alpha = 0.00843043657.ΔtHU=αTC,α=0.00843043657.
Heisenberg time for “borrowing” energy ΔE\Delta EΔE (time–energy uncertainty):

ΔtH≳ℏ2 ΔE.\Delta t_{\text{H}} \gtrsim \frac{\hbar}{2\,\Delta E}.ΔtH≳2ΔEℏ.
Two reasonable choices for ΔE\Delta EΔE:
# Full rest energy of a hydrogen atom: ΔE=mHc2\Delta E = m_H c^2ΔE=mHc2.

⇒ΔtH,rest=ℏ2mHc2=TC4π.\Rightarrow \Delta t_{\text{H,rest}}=\frac{\hbar}{2 m_H c^2}=\frac{T_C}{4\pi}.⇒ΔtH,rest=2mHc2ℏ=4πTC.
# Only the atomic binding energy: ΔE≈13.6 eV\Delta E \approx 13.6\ \text{eV}ΔE≈13.6 eV.

⇒ΔtH,bind=ℏ2 (13.6 eV).\Rightarrow \Delta t_{\text{H,bind}}=\frac{\hbar}{2\,(13.6\ \text{eV})}.⇒ΔtH,bind=2(13.6 eV)ℏ.
===== - TC≈4.4054×10−24 sT_C \approx 4.4054\times 10^{-24}\ \text{s}TC≈4.4054×10−24 s =====
* ΔtHU=αTC≈3.714×10−26 s\Delta t_{\text{HU}} = \alpha T_C \approx 3.714\times 10^{-26}\ \text{s}ΔtHU=αTC≈3.714×10−26 s
* ΔtH,rest=TC/(4π)≈3.506×10−25 s\Delta t_{\text{H,rest}} = T_C/(4\pi) \approx 3.506\times 10^{-25}\ \text{s}ΔtH,rest=TC/(4π)≈3.506×10−25 s
* ΔtH,bind≈2.420×10−17 s\Delta t_{\text{H,bind}} \approx 2.420\times 10^{-17}\ \text{s}ΔtH,bind≈2.420×10−17 s

===== - Versus full rest energy: ΔtHUΔtH,rest=4π α≈0.106.\frac{\Delta t_{\text{HU}}}{\Delta t_{\text{H,rest}}} = 4\pi\,\alpha \approx 0.106.ΔtH,restΔtHU=4πα≈0.106. Your ΔtHU\Delta t_{\text{HU}}ΔtHU is about 10.6%10.6\%10.6% of the Heisenberg time for ΔE=mHc2\Delta E=m_Hc^2ΔE=mHc2, i.e., ~9.4×9.4\times9.4× shorter. =====
* Versus 13.6 eV binding energy: ΔtHUΔtH,bind≈1.53×10−9.\frac{\Delta t_{\text{HU}}}{\Delta t_{\text{H,bind}}} \approx 1.53\times 10^{-9}.ΔtH,bindΔtHU≈1.53×10−9. Your ΔtHU\Delta t_{\text{HU}}ΔtHU is ~10910^9109 times shorter than that bound.

===== - If by “appearance of a virtual hydrogen atom” you mean an off-shell composite with energy uncertainty comparable to its full rest energy, the relevant Heisenberg scale is ΔtH,rest∼TC/(4π)\Delta t_{\text{H,rest}}\sim T_C/(4\pi)ΔtH,rest∼TC/(4π). On that basis, your coherence step time is notably smaller than the usual ℏ/(2mHc2)\hbar/(2 m_H c^2)ℏ/(2mHc2) allowance. =====
* If you only “borrow” the binding energy, the Heisenberg time is vastly larger, and your ΔtHU\Delta t_{\text{HU}}ΔtHU is far below that.

If you want, I can also express this entirely in your HU terms: since ΔtH,rest=TC/(4π)\Delta t_{\text{H,rest}} = T_C/(4\pi)ΔtH,rest=TC/(4π), your choice α\alphaα effectively sets

ΔtHU=(4πα) ΔtH,rest,\Delta t_{\text{HU}} = (4\pi\alpha)\,\Delta t_{\text{H,rest}},ΔtHU=(4πα)ΔtH,rest,
so tuning α\alphaα above/below 1/(4π)≈0.07961/(4\pi)\approx 0.07961/(4π)≈0.0796 moves you above/below the rest-energy Heisenberg scale. Your α=0.00843\alpha=0.00843α=0.00843 puts you an order of magnitude below it.